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Question: Find the last two digits of the number \({3^{600}}\)....

Find the last two digits of the number 3600{3^{600}}.

Explanation

Solution

Here, we are required to find the last two digits of the given exponential number. We will write the given number as a difference of two numbers. Then, we will use the formula of the binomial theorem to further expand the polynomial formed and solve it further to find the required last two terms of the given number.

Formula Used:
(a+b)n=nC0anb0+nC1an−1b1+nC2an−2b2+......nCna0bn{\left( {a + b} \right)^n} = {}^nC{_0}{a^n}{b^0} + {}^nC{_1}{a^{n - 1}}{b^1} + {}^nC{_2}{a^{n - 2}}{b^2} + ......{}^nC{_n}{a^0}{b^n}

Complete step by step solution:
In order to find the last two digits of the given exponential number, we will write it as:
3600=32×300=9300=(10−1)300{3^{600}} = {3^{2 \times 300}} = {9^{300}} = {\left( {10 - 1} \right)^{300}}
Now, to expand this polynomial, we will use a binomial theorem.
Now by substituting a=10a = 10, b=−1b = - 1and n=300n = 300 in the formula (a+b)n=nC0anb0+nC1an−1b1+nC2an−2b2+......nCna0bn{\left( {a + b} \right)^n} = {}^nC{_0}{a^n}{b^0} + {}^nC{_1}{a^{n - 1}}{b^1} + {}^nC{_2}{a^{n - 2}}{b^2} + ......{}^nC{_n}{a^0}{b^n}, we get,
(10−1)300=300C010300(−1)0+300C110299(−1)1+300C210298(−1)2+......300C300100(−1)300{\left( {10 - 1} \right)^{300}} = {}^{300}C{_0}{10^{300}}{\left( { - 1} \right)^0} + {}^{300}C{_1}{10^{299}}{\left( { - 1} \right)^1} + {}^{300}C{_2}{10^{298}}{\left( { - 1} \right)^2} + ......{}^{300}C{_{300}}{10^0}{\left( { - 1} \right)^{300}}
Simplifying the expression, we get
⇒(10−1)300=10300−300×10299+......+1\Rightarrow {\left( {10 - 1} \right)^{300}} = {10^{300}} - 300 \times {10^{299}} + ...... + 1
Here, all the terms are divisible by 100 except the last term.

Thus, the last two digits of the given exponential number 3600{3^{600}} will be 01.
Hence, this is the required answer.

Note:
An alternate and quick way of solving the question is:
We should know the cycle of 3, i.e.
31{3^1} ends in 3
32{3^2} ends in 9
33{3^3} ends in 7
34{3^4} ends in 1
And it repeats again and again in this manner.
Now, another property to remember is that whenever an odd number is having a power of multiple of 20, then, it always ends in 01
Whereas, whenever an even number is having a power of multiple of 20, then, it always ends in 76.
As in this question,
3600{3^{600}} can be written as 320×30{3^{20 \times 30}}
Hence, 3 is an odd number having a power which is multiple of 20.
Thus, clearly, without solving further or using binomial theorem, we can state that 3600{3^{600}} will have the last two digits as 01.
Hence, this is the required answer.